# Directional derivative of piecewise defined function?

Let $$f(x,y)=\begin{cases}\frac{x^3+y^3}{x^2-y^2},\ x^2-y^2\neq 0 \\ \ \ \ \ 0 \ \ \ \ ,x^2-y^2=0\end{cases}$$

Then find the directional derivative of $$f$$ at $$(0,0)$$ in the direction of vector $$\langle\frac45,\frac35 \rangle$$.

I don't know how to calculate the directional derivative of piecewise defined function.

How can I solve this ?

You can just use the definition $$\partial_{(4/5,3/5)}f(0,0) = \lim_{t \to 0} \dfrac{f(0+4t/5, 0 + 3t/5)-f(0,0)}{t} = \frac{13}{5}$$
Let $$v=\langle v_1,v_2\rangle=\langle\frac45,\frac35 \rangle$$. Then $$v_1^2 \ne v_2^2$$, hence the directional derivative is given by
$$\lim_{t \to 0} \frac{f(tv)-f(0,0)}{t}.$$